2/(16+6*7^(1/2))^(1/2)

pushputkc,

can you pls elaborate what you did after squaring both sides (i.e. x^2)?

Hi, let me try.

First of all, whats a 'Conjugate'? A number of the form a + √b has its conjugate as a - √b and vice versa. Conjugates are useful in certain cases as in this question.

After writing x^2 = 4/(16+6√7) .. what he has done is multiplied both numerator and denominator by the conjugate of 16+6√7, i.e., multiplied both numerator and denominator by 16-6√7.. this will help in eliminating the √ sign from the denominator, thus denominator will be 'rationalised' (rational number) and we are likely to get an answer in the form as given in the question options.
So here is what he did:

x^2 = 4/(16+6√7)

Multiply both numerator and denominator by 16-6√7..

x^2 = 4*(16-6√7)/(16+6√7)*(16-6√7)
= 4 (16-6√7)/[(16)^2- (6√7)^2] (denominator becomes (a+b)(a-b) = a^2 - b^2)
= 4 (16-6√7)/(256-252)
=4 (16-6√7)/4 = 16-6√7

So x^2 simplifies to 16-6√7

Let \(a=16\) and \(b=6\sqrt{7}\)

Substituting into the given expression, we get:
\(\frac{2}{\sqrt{16+6\sqrt{7}}} = \frac{2}{\sqrt{a+b}} = \frac{2}{\sqrt{a+b}} * \frac{\sqrt{a-b}}{\sqrt{a-b}} = \frac{2\sqrt{a-b}}{\sqrt{a^2-b^2}}\)

Since \(a=16\) and \(b=6\sqrt{7}\), we get:
\(a^2-b^2 = 16^2 - (6\sqrt{7})^2\) = 256-252 = 4

Thus:
\(\frac{2\sqrt{a-b}}{\sqrt{a^2-b^2}} = \frac{2\sqrt{16-6\sqrt{7}}}{\sqrt{4}} = \sqrt{16-6\sqrt{7}}\)

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